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I've been struggling with this for a while, but I am not smart enough to figure it out.

Suppose I have a weighted average of an economic variable $x$ across $n$ firms:

$$x=\sum_{i=1}^{n}x_i\lambda_i$$

where $\lambda_i=L_i/L$ is the employment share of firm $i$ and $L$ is total employment in the economy (sum of all firms' employment).

My question is as follows: what is the most concise way to describe a change in $x$ due to the exit of one firm from the economy. By firm exit, I mean that the total number of firms is $n-1$, and total employment is $L-L^e$ where $L^e$ is the employment of the exiting firm. Note that I am assuming that exiting employees are not reabsorbed, such that the employment levels of remaining firms remain unchanged (but not their employment shares!).

I want to get term describing change in $x$ with respect to the exit of a single firm in a simplified form such that I can describe the conditions necessary for an increase in $x$ upon firm exit.

Here's a simple example: Consider an economy with three firms:

$$x=\frac{1}{10}*x_1+\frac{3}{10}x_2+\frac{6}{10}x_3$$

Now suppose firm $3$ exits. The new value of $x$, call it $x'$ is:

$$x'=\frac{1}{4}x_1+\frac{3}{4}x_2$$

So change in $x$ is given as:

$$\Delta x = x'-x$$

What I want to obtain is a generalized form for $\Delta x$ that can then be signed based on the value of different parameters or assumptions (perhaps that the exiting firm is sufficiently small or large in terms of its employment).

Schiav
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  • This is a little unclear. To model a firm leaving, I think you just rescale the weights to the new total (perhaps that assumes the employees distribute themselves proportionally). But what is the "term" and what does "sign the difference term" mean? If you [edit] the question to include a small example perhaps we can help. Don't clarify in comments. [edit] the question. – Ethan Bolker Jun 22 '21 at 16:13
  • @EthanBolker Perhaps I have misunderstood the question. It seems to me, that if you denote the company that is leaving as having index $n$, then the new value of $x$ is simply $$\sum_{i=1}^{n-1}x_i \alpha_i.$$ Here, I define $$\alpha_i = \lambda_i \times \frac{L}{L - L^e}.$$ – user2661923 Jun 22 '21 at 17:13

1 Answers1

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With your notation, let $$ T = \lambda_1 x_1 + \cdots + \lambda_{n-1} x_{n-1} . $$ That's the total economy (in some sense) without the last firm.

Then $$ x = T + \lambda_n x_n \text{ and } x' = \frac{T}{1 - \lambda_n} $$ so $$ \begin{align} x' -x &= \frac{T}{1 - \lambda_n} -( T + \lambda_n x_n) \\ &= T\left( \frac{1}{1-\lambda_n} -1 \right)- \lambda_n x_n \\ &= T\left(\frac{ \lambda_n}{1-\lambda_n}\right) - \lambda_n x_n \\ &= \lambda_n \left( \frac{T}{1- \lambda_n} - x_n \right). \end{align} $$

That is positive just when $x_n < T/(1- \lambda_n)$ and small (in absolute value) when $\lambda_n$ is small. I think it's easy to interpret those conditions in your context.

Ethan Bolker
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